$A$ circular disc of radius $R$ meter and mass $M$ kg is rotating around the axis perpendicular to the disc. An external torque is applied to the disc such that $\theta(t) = 5t^2 - 8t$,where $\theta(t)$ is the angular position of the rotating disc as a function of time $t$. How much power is delivered by the applied torque,when $t = 2$ s (in $MR^2$)?

The $M.I.$ of a body about the given axis is $1.2 \, kg \cdot m^2$ and initially the body is at rest. In order to produce a rotational kinetic energy of $1500 \, J$,an angular acceleration of $25 \, rad/s^2$ must be applied about that axis for a duration of ........ $s$.

$A$ uniform solid cylinder of mass $M$ and radius $R$ can freely rotate around its axis $O$. There is an elastic string of relaxed length $L$ and stiffness $K$ attached to the cylinder and a static wall. Initially,the string is relaxed. As the cylinder starts rotating,the string will wind around the cylinder. The surface of the cylinder is very rough,so the string does not slip on the cylinder's surface. The minimum initial angular speed of the cylinder,${\omega _0}$,so that it can rotate through an angle $2\pi$ is (Assume Hooke's law to be valid.)

When $12000 \ J$ of work is done on a flywheel,its frequency of rotation increases from $10 \ Hz$ to $20 \ Hz$. The moment of inertia of the flywheel about its axis of rotation is (take $\pi^2=10$):

Three objects,$A$ (a solid sphere),$B$ (a thin circular disk),and $C$ (a circular ring),each have the same mass $M$ and radius $R$. They all spin with the same angular speed $\omega$ about their own symmetry axes. The amounts of work $(W)$ required to bring them to rest would satisfy the relation:

$A$ thin uniform rod $AB$ of mass $m$ and length $l$ is hinged at one end $A$ to the ground level. Initially, the rod stands vertically and is allowed to fall freely to the ground in the vertical plane. The angular velocity of the rod when its $B$ end strikes the ground is $(g = \text{acceleration due to gravity})$